Skip to content
Problems · Problem 3.5

Q.Calculate pH and pOH of 0.01 M HCl solution.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
6% · 5/79 Questions
✓ Free question

A strong monoprotic acid at 0.01 M gives [H3O+]=10−2[\mathrm{H_3O^+}] = 10^{-2} M: pH =2= 2, pOH =12= 12.

Step 1. HCl is a strong acid (section 3.4), dissociating completely, so [H3O+]=c=0.01 M=1×10−2[\mathrm{H_3O^+}] = c = 0.01\ \text{M} = 1\times10^{-2} M.

Step 2. pH=−log⁡10[H3O+]=−log⁡10(1×10−2)=2\text{pH} = -\log_{10}[\mathrm{H_3O^+}] = -\log_{10}(1\times10^{-2}) = 2.

Step 3. Using the relation pH + pOH = 14 (Eq. 3.18, at 298 K): pOH=14−pH=14−2=12\text{pOH} = 14 - \text{pH} = 14 - 2 = 12.

✓Final answer

pH =2= 2; pOH =12= 12 -- digit-for-digit the textbook's printed finals.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.